Question: Solve for $x$ : $3x^2 + 42x + 144 = 0$
Explanation: Dividing both sides by $3$ gives: $ x^2 + {14}x + {48} = 0 $ The coefficient on the $x$ term is $14$ and the constant term is $48$ , so we need to find two numbers that add up to $14$ and multiply to $48$ The two numbers $6$ and $8$ satisfy both conditions: $ {6} + {8} = {14} $ $ {6} \times {8} = {48} $ $(x + {6}) (x + {8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 6) (x + 8) = 0$ $x + 6 = 0$ or $x + 8 = 0$ Thus, $x = -6$ and $x = -8$ are the solutions.